Question

someone help :)
(a) if the function f(x)=x^3+ax^2+bx has a local minimum value -2/9sqrt(3) at x=1sqrt(3), what are the values of a and b?

(b) which of the tangent lines to the curve in part (a) has the smallest slope? @Mathematics

Answer 1

We have \(f(x)=x^3+ax^2+bx\), has a local minimum value \(\frac{-2}{9\sqrt{3}}\) at \(x=\frac{1}{\sqrt{3}}\). Are the numbers I wrote right?

Answer 2

Well, I will assume they are. What you need to know here is that when a function has a local extreme at a point \(x\), then this point has to be a critical point (ie. the derivative is equal to zero at this point, since our function is just a polynomial). I will start do this part in a new post.

Answer 3

\(f'(x)=3x^2+2ax+b \implies 3\frac{1}{3}+2a\frac{1}{\sqrt{3}}+b=0 \implies b=-\frac{2}{\sqrt{3}}a-1.\) This is only one equation in two variables, we clearly need another equation.
We will use the information the value of the function at this point is give. That's \(f(\frac{1}{\sqrt{3}})=\frac{-2}{9\sqrt{3}} \implies \frac{1}{3^{3/2}}+\frac{a}{3}+\frac{b}{\sqrt{3}}=\frac{-2}{9\sqrt{3}}\).
All you need now is to solve the two equations for the two unknowns.